Determinant by Multiplicativity: Real vs Complex

How many algebraic axioms does it take to single out the determinant among all functions on $n\times n$ matrices? Over $\mathbb C$ the answer is two; over $\mathbb R$ the same two fail, and a third — fixing the value on a single reflection — is needed. This note states both characterizations in their simplest form and explains the one structural reason for the difference.

The two axioms¶

Throughout, $f$ is a function on $n\times n$ matrices (over $\mathbb C$ or $\mathbb R$), $I$ is the identity, and $\lambda I$ is the scalar matrix. We use:

• (H1) Multiplicativity. $f(AB)=f(A)\,f(B)$ for all matrices $A,B$ (singular ones included).

• (H2) Homogeneity. $f(\lambda I)=\lambda^{\,n}$ for every scalar $\lambda\neq 0$.

Both are satisfied by the determinant, since $\det(AB)=\det A\,\det B$ and $\det(\lambda I)=\lambda^n$.

Complex: (H1) and (H2) are enough¶

Over $\mathbb C$, multiplicativity and homogeneity already force $f$ to be the determinant:

theorem eq_det_of_mul_of_scalar_pow_fin {n : ℕ} [NeZero n]
    (f : Matrix (Fin n) (Fin n) ℂ → ℂ)
    (H1 : ∀ A B, f (A * B) = f A * f B)
    (H2 : ∀ μ : ℂˣ, f (μ • 1) = μ ^ n) :
    ∀ A, f A = A.det

Here f A = A.det is det written via Mathlib’s Leibniz formula

and the theorem says: there is exactly one multiplicative, homogeneous function on complex matrices, namely $\det$.

Real: (H1) and (H2) are not enough¶

Over $\mathbb R$ the very same two axioms no longer pin $f$ down. For even $n$, the function $f(A)=|\det A|$ is a second solution: it is multiplicative ($|\det(AB)|=|\det A|\,|\det B|$) and homogeneous, because for even $n$

yet $|\det|\neq\det$ on any matrix of negative determinant. The Lean counterexample makes this precise:

theorem not_characterization_real_even [Nonempty n] (hn : Even (Fintype.card n)) :
    ¬ ∀ (f : Matrix n n ℝ →* ℝ),
        (∀ μ : ℝˣ, f (μ • 1) = (μ : ℝ) ^ Fintype.card n) → ∀ A, f A = A.det

The fix is one extra normalization on a reflection $T=\operatorname{diag}(-1,1,\dots,1)$, which has $\det T=-1$:

• (H3) Reflection. $f(T)=-1$.

This excludes $|\det|$ at a stroke (since $|\det T|=1\neq -1$) and restores uniqueness for every $n$:

theorem eq_det_of_mul_of_scalar_pow_of_reflection_fin {n : ℕ} [NeZero n]
    (f : Matrix (Fin n) (Fin n) ℝ → ℝ)
    (H1 : ∀ A B, f (A * B) = f A * f B)
    (H2 : ∀ μ : ℝˣ, f (μ • 1) = μ ^ n)
    (H3 : f (Matrix.diagonal (fun i => if i = 0 then -1 else 1)) = -1) :
    ∀ A, f A = A.det

For odd $n$, (H3) is automatic — it adds nothing — so it is a clean, uniform hypothesis.

Definition versus characterization¶

It is tempting to object: the determinant is the Leibniz polynomial $L(A)=\sum_\sigma\operatorname{sgn}(\sigma)\prod_i A_{i,\sigma(i)}$, one explicit formula valid over any commutative ring. We proved it exists and is unique over $\mathbb C$; the very same formula computes the determinant of a real matrix. So why isn’t the complex result enough for real matrices, with $|\det|$ “excluded automatically” by the complex numbers?

The resolution is to separate two different roles the word determinant is playing.

• Definition (transfers). $L$ is a single formula. Defining $\det:=L$ works over $\mathbb R$ as the restriction of the complex $\det$: for a real matrix $A$, $\det_{\mathbb R}(A)=\det_{\mathbb C}(A)$. The $|\det|$ issue never touches the definition.

• Characterization (does not transfer). “$\det$ is the unique (H1)+(H2) function” is a statement about the space of functions over a given field — and that space is bigger over $\mathbb R$ (it contains $|\det|$ for even $n$) than over $\mathbb C$ (just $\det$).

The reason the complex characterization cannot be “imported” to the reals is that a function on real matrices is not the restriction of a function on complex matrices. Given a real $f$ with (H1)+(H2), there is no canonical complex extension, so the complex uniqueness theorem does not apply to it. Concretely, $|\det|$ is invisible to the complex world:

• every complex (H1)+(H2) function is $\det_{\mathbb C}$, whose restriction to real matrices is $\det_{\mathbb R}$ — not $|\det|$;

• $|\det|$ does extend to $|\det_{\mathbb C}|$, but that fails complex (H2): $|\det_{\mathbb C}(\mu I)|=|\mu|^n\neq\mu^n$ for a general complex $\mu$. It slips past real (H2) only because a real $\mu^n\ge 0$ when $n$ is even.

In other words, real (H2) is the same equation quantified over fewer scalars, and $|\det|$ lives precisely in that gap.

The one structural reason: $\mathbb C^\times$ is divisible, $\mathbb R^\times$ is not¶

Both characterizations reduce to the same final step. The factorization theorem gives $f(A)=g(\det A)$ for a homomorphism $g:K^\times\to K^\times$ read off the diagonal, and (H2) forces $g(u)^n=u^n$, i.e.

Uniqueness is exactly the statement $\varepsilon\equiv 1$ (then $g=\operatorname{id}$ and $f=\det$). Now a homomorphism from a divisible group into a finite group is automatically trivial: if $w=z^n$ then $\varepsilon(w)=\varepsilon(z)^n=1$. So everything hinges on whether $K^\times$ is divisible — i.e. whether every element has an $n$-th root.

• $\mathbb C^\times$ is divisible — every nonzero complex number has an $n$-th root (this is algebraic closure). Hence $\varepsilon\equiv 1$ for free, and (H1)+(H2) suffice.

• $\mathbb R^\times\cong\{\pm1\}\times\mathbb R_{>0}$ is not divisible. The factor $\mathbb R_{>0}$ is divisible (so $\varepsilon=1$ there), but $\{\pm1\}$ is not: for even $n$, $\mu_n(\mathbb R)=\{\pm1\}$ and $\varepsilon(-1)$ may equal -1, giving $g=\operatorname{sign}$ and $f=|\det|$. (H3) forces $\varepsilon(-1)=1$, removing the last bit of freedom.

So the gap between the real and complex characterizations is not about the determinant formula at all — it is the failure of $\mathbb R$ to contain enough roots. The determinant is the same object over both fields; only the question “which axioms pin it down” is field-sensitive, and the missing roots of $\mathbb R$ are paid for by exactly one reflection.